# 4.1.15. Thermodynamics¶

## 4.1.15.1. What are examples of path functions and state variables?¶

• Path functions: Heat and work (about the boundary)

• State variables: Internal energy, enthalpy, entropy (about the fluid)

## 4.1.15.2. What is the First Law of Thermodynamics and what kind of processes is it applicable to?¶

$\text{Net heat added to system} + \text{Net work done on system} = \text{Increase in internal energy of system}$
$dq + dw = de$

(lower case implies per unit mass)

The first law applies to reversible and irreversible processes

## 4.1.15.3. What is the Second Law of Thermodynamics and what is it for a reversible process?¶

$\text{Gross heat added to system} + \text{Net work done by system} \ge 0$
$dq + \Sigma = Tds$

i.e. there is irreversible work done by internal molecular changes $$\Sigma$$

For a reversible process $$\Sigma = 0$$

i.e. $$ds = {dq \over T}$$

## 4.1.15.4. What is Enthalpy?¶

From the $$1^{st}$$ law, for both reversible and irreversible processes:

$dq = d(e+pv) = dh$

where $$dw = -d(pv)$$

Enthalpy is a state variables

For both reversible and irreversible processes it equals the internal energy plus pressure-volume potential energy.

## 4.1.15.5. What is Entropy?¶

From the $$2^{nd}$$ law for a reversible process:

$s_2 - s_1 = \int_1^2 {dq \over T}$

Entropy is a macro state variable

For a reversible process it is the infinitisimal heat added over constant temperature (non-adiabatic).

## 4.1.15.6. What are the two property relations for entropy change from the 1st and 2nd laws of thermodynamics at constant pressure and constant volume?¶

• 1st Law (reversible and irreversible) $$\longrightarrow de = dw+dq$$

• Reversible work done by system $$\longrightarrow dw = -pdv$$

• Reversible heat added to system (2nd law) $$\longrightarrow dq = Tds$$

• For reversible and irreversible processes (1) $$\longrightarrow Tds = de + pdv$$

• Definition of enthalpy $$\longrightarrow dh = d(e+pv) = de + d(pv) = de + pdv + vdp$$

• By rearrangement $$\longrightarrow de = dh - pdv - vdp$$

• For reversible and irreversible processes (2) $$\longrightarrow Tds = dh - vdp$$

## 4.1.15.7. What is the entropy change for an ideal gas for a reversible and irreversible process?¶

• $$ds = {de \over T}+{p \over T}dv$$

• $$ds = {dh \over T} - {v \over T}dp$$

• Calorically perfect

$$\longrightarrow de=c_vdT$$

$$\longrightarrow dh=c_pdT$$

• Ideal gas

$$\longrightarrow p = {RT \over v}$$

$$\longrightarrow v = {RT \over p}$$

• Hence

$$\longrightarrow s_2 - s_1 = c_v \int_1^2 {dT \over T} + R \int_1^2 {dv \over v} = c_v ln{T_2 \over T_1} + R ln{v_2 \over v_1} = c_v ln{T_2 \over T_1} + R ln{\rho_1 \over \rho_2}$$

$$\longrightarrow s_2 - s_1 = c_p \int_1^2 {dT \over T} - R \int_1^2 {dp \over p} = c_p ln{T_2 \over T_1} - R ln{p_2 \over p_1}$$

## 4.1.15.8. How are the isentropic relations for an ideal gas derived from the entropy change?¶

• Isentropic $$\longrightarrow s_2 - s_1 = 0$$

• Hence

$$\longrightarrow 0 = c_v ln{T_2 \over T_1} - R ln{\rho_2 \over \rho_1}$$

$$\longrightarrow 0 = c_p ln{T_2 \over T_1} - R ln{p_2 \over p_1}$$

• Definition of $$c_v$$ and $$c_p$$:

$$c_v = {R \over {\gamma -1}}$$

$$c_v = {{\gamma R} \over {\gamma -1}}$$

• Hence:

$$\longrightarrow ln{\rho_2 \over \rho_1} = {1 \over {\gamma -1}} ln{T_2 \over T_1}$$

$$\longrightarrow ln{p_2 \over p_1} = {{\gamma} \over {\gamma -1}} ln{T_2 \over T_1}$$

• Hence:

$$\longrightarrow {\rho_2 \over \rho_1} = {T_2 \over T_1}^{1 \over {\gamma -1}}$$

$$\longrightarrow {p_2 \over p_1} = {T_2 \over T_1}^{{\gamma} \over {\gamma -1}}$$

• Hence:

$$\longrightarrow {p_2 \over p_1} = {\rho_1 \over \rho_2}^{\gamma}$$

$$\longrightarrow {p \over {\rho^{\gamma}}} = \alpha$$